Problem: Two reals $a$ and $b$ are such that $a+b=7$ and $a^3+b^3=91$. Compute $ab$.
Solution: We have $91=a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)((a+b)^2-3ab)=7\cdot (49-3ab)$, from which $ab=\boxed{12}$.